Graph this system of equations and solve. $10x+6y = -12$ $4x-6y = -30$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Answer: Convert the first equation, $10x+6y = -12$ , to slope-intercept form. $y = -\dfrac{5}{3} x - 2$ The y-intercept for the first equation is $-2$ , so the first line must pass through the point $(0, -2)$ The slope for the first equation is $-\dfrac{5}{3}$ . Remember that the slope tells you rise over run. So in this case for every $5$ positions you move down (because it's negative) You must also move $3$ positions to the right. $3$ positions to the right. $5$ positions down from $(0, -2)$ is $(3, -7)$ Graph the blue line so it passes through $(0, -2)$ and $(3, -7)$ Convert the second equation, $4x-6y = -30$ , to slope-intercept form. $y = \dfrac{2}{3} x + 5$ The y-intercept for the second equation is $5$ , so the second line must pass through the point $(0, 5)$ The slope for the second equation is $\dfrac{2}{3}$ . Remember that the slope tells you rise over run. So in this case for every $2$ positions you move up You must also move $3$ positions to the right. $3$ positions to the right. $2$ positions up from $(0, 5)$ is $(3, 7)$ Graph the green line so it passes through $(0, 5)$ and $(3, 7)$ The solution is the point where the two lines intersect. The lines intersect at $(-3, 3)$.